Calculating $\lim_{x \to 1} \frac{\sqrt[3]{x}-1}{x-1}$ without L´Hôpital rule.

Question

How to calculate the limit above without using L´Hôpital rule? I am not able to factor the numerator and I´ve also tried factoring the numerator.

Answer 1

HINT

$$a^{3} - b^{3} = (a-b)(a^{2} + ab + b^{2})$$

Answer 2

Nothing new, but little modification to previous answers for your easier understanding. Put $\sqrt[3]{x}=y.$ Then your problem is $$\lim_{y\rightarrow 1}\frac{y-1}{y^3-1}= \lim_{y\rightarrow 1}\frac{y-1}{(y-1)(y^2+y+1)}=\frac{1}{3}.$$